PERCENTAGES
Are represented by quantities compared to 100 units, which is considered a whole
Example 1
25 cars are red and 75 are white out of 100 cars in a dealer parking lot.
We say 25 cars per 100 are red, or 25 cars / 100 are read or 25% of the cars are red.
Example 2
The bank is offering $ 0.25 for each $100 deposited in a checking account per year.
We could say that it offers $0.25 per 100, or $0.25 / 100 or 0.25%.
Mathematically we could represent a percentage as 12% 
This expression could be considered A% 
B
, so if we know 2 variables out of 3 we
C
could calculate the un-known
The 3 types of exercises with percentages
1. How much is 24% out of 201?
B=A% x C = 24% x 201 
24
x 201  48.24
100
2. If 15% of a quantity is 30, what is the initial quantity C?
B
30
30
100
so C 

 30 x
 200
15
A%
15%
15
100
3. What percentage is 25 out of 600?
C=
A% 
25
 0.0416  4.16%
600
12
100
Exercise types
We mix 2 same chemicals of 2 different concentrations. One chemical has the
concentration of 25% (solution A) and the other one has the concentration of 35%
(solution B).
How many litres of each solution we need to mix to obtain 100 litres of solution with
30% concentration (Solution C)?
The equations we could write with the data given by the exercise are:
a+b=100 (1)
a=quantity in litres of 25% concentration; b= quantity in litres of 32% concentration
In other words if we mix 5 litres of solution A with 7 litres of Solution B we will get 12
litres of the new solution.
The quantity of pure chemical in the final solution is the same with the pure chemical of
the 2 initial chemicals we are mixing.
Quant. pure chem. in solution A+ Quant. pure chem.. in solution B= Quant. pure
chemical in solution C
(2)
So the quantity of pure chemical in solution A is:
a x 25%
(3)
The quantity of pure chemical in solution B is:
b x 35%
(4)
The quantity of pure chemical in solution C is:
100 litres x 30% (5)
Substituting in equation (2) the values calculated we get :
a x 25% + b x 35% = 100 x 30% (6)
We have a set (system) of 2 equations with 2 variables to solve (1) and (6):
a+b=100
a x 25% + b x 35% = 100 x 30% or we re-write :
a+b=100
0.25a + 0.35b = 0.3 x 100 (7)
We will solve it by elimination.
We multiply left and right the equation (1) with – 0.25, than we add it to ( 7)
- 0.25 a – 0.25 b = - 0.25 x 100
0.25 a + 0.35 b = 0.3 x 100
--------------------------------------0.10 b = 5 => b = 5/0.10 = 50 litres
So a = 100 – b = 100 - 50 = 50 litres
A store decided to sell a mixture of sun flower seeds with pumpkin seeds
The cost of the sun flower seeds is $2/lbs and the pumpkin seeds are sold at
$3/lbs.
What quantity of each flavour should be taken if the store manager decided
to sell 20lbs of mixture with $2.75/lbs.
Weuse the same logic as in the example above
a+b=20  represents the total weight obtained by adding the 2 type of seeds
2a+3b=20x2.75 the cost of the mixture, by adding the 2 type of seeds
-2a-2b=-40  we use the elimination method
---------------b=55-40=15 lbs
a=20-15=5 lbs